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Please for the substition method solving $\int\frac{dx}{\cos^6 x}$

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$\int\frac{dx}{\cos^6 x}$=$\int\frac{1}{\cos^4 x}\cdot\frac{dx}{\cos^2 x}$=$\int\frac{1}{(\frac{1}{\sqrt{1+\tan^2 x}})^4}\cdot\frac{dx}{\cos^2 x}$=$\int(1+\tan^2 x)\cdot\frac{dx}{\cos^2x}$=$|\tan x=t\Rightarrow\frac{dx}{\cos^2 x}=dt|$=$\int(1+t^2)^2dt$=$\int(1+2t^2+t^4)dt$=$t+\frac{2}{3}t^3+\frac{1}{5}t^5$=$\tan x+\frac{2}{3}\tan^3 x+\frac{1}{5}\tan^5 x+C$.

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HINT:

$$\frac1{\cos^6 x}=\sec^6 x=\left(\sec^2x\right)^2\sec^2x=\left(1+\tan^2x\right)^2\sec^2x$$

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